Transition Matrix from B to C

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If $B=\{ 2+x,1+2x\}$ and $C=\{ 1+x, 1-x\}$ are 2 basis for $P_1$, and $v=-3x+4$ find $[v]_B$, $_BP_C$ and $[v]_C$.

my attempt: Since $B$ is a basis for $V$, then any $v\in B$ can be written "uniquely" as a linear combination of the vectors of $B$.

then $v=-3x+4$ can be written uniquely in the form:

$-3x+4=c_1(2+x)+c_2(1+2x)$ where $c_1$ and $c_2$ are scalars.

$-3x+4=c_1(2+x)+c_2(1+2x)$

$=2c_1+c_2+(c_1+2c_2)x$

Then

$2c_1+c_2=4$

$c_1+2c_2=-3$

and by solving this system, we have

$c_1=\frac{11}{3}$ and $c_2=\frac{-10}{3}$

so, $[v]_B=$ $\left[ {\begin{array}{c} \frac{11}{3} \\ \frac{-10}{3} \end{array} } \right] $

The transition matrix from $B$ to $C$ is

$_BP_C=$ $\left[ {\begin{array}{cc} [2+x]_C & [1+2x]_C \end{array} } \right] $

we can find $[2+x]_c$ as follows

$2+x=\alpha_1(1+x)+\alpha_2(1-x)$

$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$

and hence

$\alpha_1+\alpha_2=2$

$\alpha_1-\alpha_2=1$

and by solving this system we get that

$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{1}{2}$

Similarly,

$1+2x=\alpha_1(1+x)+\alpha_2(1-x)$

$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$

and hence

$\alpha_1+\alpha_2=1$

$\alpha_1-\alpha_2=2$

and by solving this system we get that

$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{-1}{2}$

$_CP_B=\left[ {\begin{array}{cc} \frac{3}{2} & \frac{3}{2} \\ \frac{-1}{2} & \frac{1}{2} \end{array} } \right] $

or,$_CP_B=\left[ {\begin{array}{cc} \frac{3}{2} & \frac{3}{2} \\ \frac{1}{2} & \frac{-1}{2} \end{array} } \right] $

what is the true?

and we know that

$[v]_C=_BP_C[v]_B$

$[v]_C=\left[ {\begin{array}{cc} \frac{3}{2} & \frac{3}{2} \\ \frac{1}{2} & \frac{-1}{2} \end{array} } \right]$$\left[ {\begin{array}{c} \frac{11}{3} \\ \frac{-10}{3} \end{array} } \right] $

$= \left[ {\begin{array}{c} \frac{1}{2} \\ \frac{7}{2} \end{array} } \right] $

and if I use $_CP_B=\left[ {\begin{array}{cc} \frac{3}{2} & \frac{3}{2} \\ \frac{-1}{2} & \frac{1}{2} \end{array} } \right] $

I got that$= \left[ {\begin{array}{c} \frac{1}{2} \\ \frac{7}{2} \end{array} } \right] $

what is the true please? Thanks.

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1 Answer

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$b_1 = 2+x = a_{11} c_1 + a_{21} c_2\\ 2+x = a_{11}(1+x) + a_{21}(1-x)\\ a_{11}+a_{21} = 2\\ a_{11}-a_{21} = 1\\ a_{11} = \frac 32, a_{21} = \frac 12$

and

$b_2 = 1+2x = a_{12} c_1 + a_{22} c_2\\ a_{12} + a_{22} = 1\\ a_{12} - a_{22} = 2\\ a_{12} = \frac 32, a_{22} = - \frac 12$

$_CP_B = \begin{bmatrix} \frac 32 &\frac 32\\\frac 12&-\frac 12\end {bmatrix}$

Here is a different way to think about it.

$B = \begin {bmatrix} 2&1\\1&2 \end{bmatrix}$

Will transform vectors represented in the basis $B$ into the standard basis.

$C = \begin {bmatrix} 1&1\\1&-1 \end{bmatrix}$

Will transform vectors represented in the basis $C$ into the standard basis.

$C^{-1} = \begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}$

Will reverse that and take vectors in the standard basis to the basis $C.$

$C^{-1}B$ will take a vector in $B$, translated to the standard basis and then transform that into the basis $C.$

$\begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}\begin {bmatrix} 2&1\\1&2 \end{bmatrix}= \begin {bmatrix} \frac 32 & \frac 32 \\ \frac 12 & -\frac 12 \end{bmatrix}$

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